Question

How do you graph `y + 1 = 3 cos 4 (x-2)`?

Answer 1

Write the equation as `y=3cos(4x-8)+1`. Now you're in the generic form

`Acos(omega x + phi)+k`.

A function in this form has four important informations:

1. `A` is the amplitude, which is the maximum value reached by the function. Of course, the standard amplitude is `1`, since `cos(x)` ranges between `-1` and `1`. And in fact, a function with amplitude `A` ranges from `-A` to `A`.

2. `omega` affects the period, because it changes the "speed" with which the function grows. Look at this example: if we have the standard function `cos(x)`, if you want to go from `cos(0)` to `cos(2\pi)`, the variable `x` must from from `0` to `2\pi`. Now try `cos(2x)`: in this case, if `x` runs from `0` to `pi`, your function ranges from `cos(0)` to `cos(2\pi)`. So, we needed "half" the `x` travel to cover a whole period. In general, the formula states that the period `T` is `T=(2\pi)/\omega`.

3. `\phi` is a phase shift, and again look at this example: with the standard function `cos(x)`, you have `cos(0)` for `x=0`, of course. Now we try `cos(x-1)`. To have `cos(0)`, we must input `x=1`. So, the same value has been shifted ahead of `1` unit. In general, if `phi` is positive, it shifts the function backwards (which means to the left on the `x`-axis) of `phi` units, and if `phi` is negative, the shift is to the right.

4. Finally, the `+1` at the end is a vertical shift. Think of it like this: when you have `y=cos(x)`, it means that you are associating with every `x` the `y` value "`cos(x)`". Now, you change to `y=cos(x)+1`. This means that now you associate to the same old `x` the new value `cos(x)+1`, which is one more than the old value. So, if you add one unit on the `y` axis, you shift upwards. Of course, if `k` is negative, the shift is downwards.

So, in the end, you start from the standard cosine function. Then, you do all the transformations:

1. Change the amplitude

2. Change the period

3. Horizontal shift

4. Vertical shift